Lesson 12: Integrals Preview: Area Under a Curve

The tangent problem asked for a slope; the area problem asks for the area under a curve, where no rectangle or triangle formula applies. The strategy mirrors everything you have seen: approximate with shapes you *can* compute (rectangles), then take a limit as the approximation refines. That limit is the definite integral — the second pillar of calculus, previewed here.

Riemann sums: rectangles under a curve

Divide [a, b] into n subintervals of width Δx = (b − a)/n and erect rectangles whose heights are function values at chosen points of each subinterval. Three standard choices: left endpoints, right endpoints, midpoints. Worked example: f(x) = x^2 on [0, 3] with n = 3, so Δx = 1:

Left sum:   1 × (f(0) + f(1) + f(2)) = 0 + 1 + 4 = 5
Right sum:  1 × (f(1) + f(2) + f(3)) = 1 + 4 + 9 = 14
Trapezoid:  (5 + 14)/2 = 9.5   (average of the two, equivalently trapezoids on each strip)

For an increasing function the left sum underestimates and the right sum overestimates — the true area is trapped between 5 and 14, and the trapezoid 9.5 is already close. Refining (n = 6, 12, 100...) squeezes both bounds onto the same number: that number is the integral. For x^2 on [0, 3] the exact value is 9 (see below), so the trapezoid rule with just three strips is within half a unit.

The definite integral as a limit

The definite integral is defined as the limit of these sums as n → ∞ (equivalently Δx → 0):

∫(a to b) f(x) dx = lim(n→∞) Σ f(x_i) Δx

The notation is a stretched S for "sum," and it records the recipe: sample heights f(x_i), multiply by widths Δx, add, take the limit. Worked example — exactness for a line. f(x) = x + 1 on [0, 4] with n = 4 midpoint rectangles: midpoints 0.5, 1.5, 2.5, 3.5 give heights 1.5, 2.5, 3.5, 4.5, sum × Δx=1 → 12. The exact area (a trapezoid under the line: (1 + 5)/2 × 4 = 12) agrees perfectly — midpoints are exact for linear functions, one reason they are favoured in practice. Recognising a limit of sums *as* an integral runs both ways: lim(n→∞) Σ (1/n)·e^(i/n) is ∫(0 to 1) e^x dx, because Δx = 1/n and the sample points are x_i = i/n.

The Fundamental Theorem preview: antiderivatives

Computing limits of sums directly is painful; the miracle of calculus is that areas are computed by antiderivatives: if F' = f, then ∫(a to b) f(x) dx = F(b) − F(a). Worked examples. ∫(1 to 3) 3x^2 dx: an antiderivative of 3x^2 is x^3, so the value is 3^3 − 1^3 = 27 − 1 = 26. ∫(0 to 3) x^2 dx: antiderivative x^3/3, value 27/3 − 0 = 9 — confirming the exact area the rectangles were hunting above. ∫(0 to 2) x^2 dx = 8/3 ≈ 2.667. The power pattern (antiderivative of x^p is x^(p+1)/(p + 1)) previews the power rule for integration and is the mirror image of the derivative power rule from Lesson 11: differentiation and integration undo each other.

Modelling: distance from a velocity table

A cyclist's speed is recorded every second: t = 0..4 s, v = 0, 6, 9, 11, 12 m/s. Distance travelled is the area under the velocity graph — approximate it with trapezoids (Δt = 1):

(0+6)/2 + (6+9)/2 + (9+11)/2 + (11+12)/2 = 3 + 7.5 + 10 + 11.5 = 32 m

Each trapezoid averages the speeds at the ends of its second and multiplies by the time — exactly the trapezoidal rule, and exactly how real instruments (GPS units, heart-rate straps, smart meters) accumulate totals from sampled data. Left and right sums give 26 and 38: the truth is bracketed, and the trapezoid sits in between.

Signs and geometry

Area under the x-axis counts negatively in an integral: ∫(a to b) f(x) dx is *signed* area. So ∫(0 to 3) x dx = 9/2 is the area of a triangle (base 3, height 3 — antiderivative x^2/2 gives 9/2 ✓ two routes, one answer), while a function dipping below the axis subtracts. Displacement from velocity is signed (net change of position); total distance travelled requires |v| — the first of many places where the integral's sign convention carries physical meaning.

Common mistakes

• Using the wrong endpoints. A left sum on [0, 3] with n = 3 uses x = 0, 1, 2 — not 1, 2, 3 (that is the right sum). Off-by-one in the sample points is the Riemann-sum equivalent of the series term-count error.
• Forgetting Δx. The sum of heights alone (0 + 1 + 4 = 5) equals the left sum here only because Δx = 1. With n = 6, Δx = 1/2 and omitting it doubles the answer.
• Assuming trapezoidal estimates are always underestimates. For concave-up curves trapezoids overestimate (9.5 vs the exact 9 above); for concave-down they underestimate. The bias flips with concavity — know which case you are in.
• Confusing the antiderivative evaluation order. F(b) − F(a), in that order: 27 − 1 = 26, not −26. A negative "area" for a positive function is a certain sign of swapped endpoints.
• Dropping the sign convention. ∫ over a region where f < 0 returns a negative number. If a problem asks for *area*, take absolute values region by region; if it asks for the integral, report the signed value.

Exam Experience

1. Using trapezoids with the table t: 0, 1, 2 and v: 5, 7, 9, the distance estimate is

• (A) 12
• (B) 14
• (C) 16
• (D) 21

Answer: (B). (5 + 7)/2 + (7 + 9)/2 = 6 + 8 = 14. (A) is the left sum (5 + 7) and (C) the right sum (7 + 9) — the trapezoid average of the two appears as the trap pair.

2. ∫(0 to 3) x dx =

• (A) 3
• (B) 9/2
• (C) 9
• (D) 6

Answer: (B). Geometrically the region is a triangle of base 3 and height 3: (3 × 3)/2 = 9/2. The antiderivative x^2/2 evaluated 0 to 3 gives the same 9/2 — (C) forgets the division by 2 in both methods.

3. The limit lim(n→∞) Σ_{i=1}^{n} (1/n)·(1 + i/n) equals

• (A) ∫(0 to 1) (1 + x) dx = 3/2
• (B) ∫(0 to 1) x dx = 1/2
• (C) ∫(1 to 2) (1 + x) dx
• (D) 1

Answer: (A). Δx = 1/n and the sample points are x_i = 1 + i/n... read carefully: with x_i = i/n, the summand (1/n)(1 + i/n) is f(x_i)Δx with f(x) = 1 + x on [0, 1], giving ∫(0 to 1) (1 + x) dx = [x + x^2/2] = 1 + 1/2 = 3/2. (C) misreads the interval: the samples i/n lie in [0, 1], so the "1 +" belongs to the function, not the interval.

Constructed response. Let f(x) = x^2 on [0, 3]. (a) Compute the left-endpoint and right-endpoint Riemann sums with n = 3. (b) Compute the trapezoidal estimate with n = 3. (c) Find the exact value of ∫(0 to 3) x^2 dx using an antiderivative. (d) Which of the three estimates is closest, and what feature of the graph explains the direction of the trapezoidal error?

Scoring guide. (a) 2 points: left sum 1 × (0 + 1 + 4) = 5 — 1 point; right sum 1 × (1 + 4 + 9) = 14 — 1 point. (b) 1 point: (5 + 14)/2 = 9.5. (c) 1 point: antiderivative x^3/3, so 27/3 − 0 = 9. (d) 2 points: the trapezoid (error 0.5) is far closer than left (error 4) or right (error 5) — 1 point; x^2 is concave up, so each trapezoid's top edge lies above the curve, making the trapezoidal rule *over*estimate — 1 point.

Going deeper: midpoint sums and the properties of integrals

Midpoint sums deserve their numbers for x^2 on [0, 3], n = 3: midpoints 0.5, 1.5, 2.5 give heights 0.25, 2.25, 6.25, sum × Δx = 8.75. Compare the field: left 5, right 14, trapezoid 9.5, midpoint 8.75, exact 9. The midpoint rule — with the same three rectangles — beats the trapezoid, and for concave-up curves it underestimates (the tangent-at-midpoint rectangle slips under the curve as much as it pokes above). Knowing the *direction* of each method's bias for a given concavity is worth more in practice than a fourth decimal place.

Properties that compute. ∫(a to b) c dx = c(b − a) (a rectangle — the integral of a constant is geometry, not calculus). Additivity over intervals: ∫(0 to 3) x^2 dx = ∫(0 to 1) x^2 dx + ∫(1 to 3) x^2 dx = 1/3 + 26/3 = 27/3 = 9 ✓ — splitting at convenient points is legal and often strategic. Average value: the average of f on [a, b] is (1/(b − a))∫ f, so x^2 averages 9/3 = 3 on [0, 3] — the height of the rectangle with the same area, and the precise meaning of "average speed" when speed varies: total distance over total time. Net change: ∫(a to b) (rate) dt = (total change) — distance from velocity, charge from current, population change from net growth rate. This theorem is why integrals appear everywhere rates are known and totals are wanted.

Workshop: displacement versus distance

A particle moves with velocity v(t) = t − 2 m/s on [0, 4]. The displacement is ∫(0 to 4) (t − 2) dt = [t^2/2 − 2t] = (8 − 8) − 0 = 0: the particle ends where it started, having gone backwards (v < 0) for t < 2 and forwards for t > 2. The total distance counts both legs positively: the velocity graph makes two triangles, each of base 2 and height 2, so distance = 2 + 2 = 4 m. The integral of v gives displacement; the integral of |v| gives distance — the signed-area convention of the main lesson, now with physical stakes. Whenever a problem says "how far did it travel in total," integrate the speed; when it says "where did it end up," integrate the velocity.